What is ((1+i)/(1-i))^{-1} ?

The solution to ((1+i)/(1-i))^{-1} is -i

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simplify ((1+i)/(1-i))^{-1}

simplify

(\frac{1 + i}{1 - i})^{- 1}

- i

Solution steps

(\frac{1 + i}{1 - i})^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(\frac{1 + i}{1 - i})^{- 1} = \frac{1}{\frac{1 + i}{1 - i}}

= \frac{1}{\frac{1 + i}{1 - i}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{1 - i}{1 + i}

Multiply by the conjugate

\frac{1 - i}{1 - i}

= \frac{( 1 - i ) ( 1 - i )}{( 1 + i ) ( 1 - i )}

Simplify

\frac{( 1 - i ) ( 1 - i )}{( 1 + i ) ( 1 - i )} : \frac{- 2 i}{2}

= \frac{- 2 i}{2}

Cancel the common factor:

2

= - i

8 x - 2

\frac{5}{7} \div \frac{5}{3}

8 - (- 1)

\frac{27}{30}

(x + 5)^{2} - (x - 5)^{2}